Question

Evaluate: ${\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx$.

• A. $\frac{\pi }{2}(\beta -\alpha )$
• B. $\frac{\pi }{2}(\alpha -\beta )$
• C. $\pi (\beta -\alpha )$
• D. $\pi (\alpha -\beta )$

Answer 1

The correct option is A $\frac{\pi }{2}(\beta -\alpha )$

Let $I={\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx$ (i)
Put $x=\alpha {\cos }^{2}t+\beta {\sin }^{2}t$
$\therefore x-\alpha =\alpha {\cos }^{2}t+\beta {\sin }^{2}t-\alpha =\beta ({\sin }^2\left(t\right)+\alpha ({\cos }^{2}t-1\left)\right)$
$=\beta {\sin }^{2}t-\alpha {\sin }^{2}t⟹(x-\alpha )=(\beta -\alpha ){\sin }^{2}t$
Similarly, $\beta -x=(\beta -\alpha ){\cos }^{2}t$
$\therefore dx=2(\beta -\alpha )\sin t\cos tdt$
where, $x=\alpha ⟹{\sin }^{2}t=0\therefore t=0$ ($\because \alpha <\beta$)
and $x=\beta ⟹{\cos }^{2}t=0\therefore t=\frac{\pi }{2}$
$\therefore$ equation (i) reduces to;
$I={\int }_0^{\frac{\pi }{2}}\sqrt{\frac{(\beta -\alpha ){\sin }^{2}t}{(\beta -\alpha ){\cos }^{2}t}}.2(\beta -\alpha )\sin t\cos tdt$
$={\int }_0^{\frac{\pi }{2}}\frac{\sin t}{\cos t}.2(\beta -\alpha )\sin t\cos tdt$
$[\because \sqrt{\frac{{\sin }^{2}t}{{\cos }^{2}t}}=+\frac{\sin t}{\cos t}astϵ[0,\frac{\pi }{2}]]$
$=(\beta -\alpha ){\int }_0^{\frac{\pi }{2}}2{\sin }^{2}tdt$
$=(\beta -\alpha ){\int }_0^{\frac{\pi }{2}}(1-\cos 2t)dt$
$=(\beta -\alpha ){(t-\frac{\sin 2t}{t})}_0^{\frac{\pi }{2}}$
$=(\beta -\alpha )[(\frac{\pi }{2}-\frac{1}{2}\sin \pi )-(0-\frac{1}{2}\sin 0)]$
$=(\beta -\alpha )\left[\frac{\pi }{2}\right]$
$\therefore I=\frac{\pi }{2}(\beta -\alpha )$
Aliter $I={\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx$
[Put $\sqrt{\beta -x}=t\therefore \beta -x=t^2$, $-dx=2tdt$]
when $x=\alpha ;x=\beta$
$\beta -\alpha =t^2;0=t^2$
$\sqrt{\beta -\alpha }=t;0=t$
$\therefore I={\int }_{\sqrt{\beta -\alpha }}^0\frac{\sqrt{(\beta -t^2)-\alpha }}{t}(-2t)dt$
$=-2{\int }_{\sqrt{\beta -\alpha }}^0\sqrt{(\beta -\alpha )-t^2}dt$
$I=2{\int }_0^{\sqrt{\beta -\alpha }}\sqrt{{\left(\sqrt{\beta -\alpha }\right)}^2-t^2}dt$
$=2{\{\frac{t}{2}\sqrt{(\beta -\alpha )-t^2}+\frac{(\beta -\alpha )}{2}{\sin }^{-1}\left(\frac{t}{\sqrt{\beta -\alpha }}\right)\}}_0^{\sqrt{\beta -\alpha }}$
$I=2[\{0+\frac{(\beta -\alpha )}{2}{\sin }^{-1}\left(1\right)\}-\{0+\frac{(\beta -\alpha )}{2}{\sin }^{-1}\left(0\right)\}]$
$=\frac{2(\beta -\alpha )}{2}.\frac{\pi }{2}$
$\therefore I=\frac{\pi }{2}(\beta -\alpha )$